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If ‘n’ is a natural number, then 4n – 3n ends with a digit x. The number of possible values of x is

Question

If ‘n’ is a natural number, then 4n – 3n ends with a digit x. The number of possible values of x is

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Solution

The problem is asking for the number of possible last digits (units place) of the expression 4n - 3n for any natural number n.

Step 1: Let's first understand the pattern of the last digit of powers of 4 and 3.

Powers of 4: 4^1 = 4, 4^2 = 16, 4^3 = 64, 4^4 = 256, 4^5 = 1024, ... The last digit repeats every two powers: 4, 6, 4, 6, ...

Powers of 3: 3^1 = 3, 3^2 = 9, 3^3 = 27, 3^4 = 81, 3^5 = 243, ... The last digit repeats every four powers: 3, 9, 7, 1, 3, 9, ...

Step 2: Now, let's subtract the last digits of powers of 3 from the last digits of powers of 4.

For n = 1, 4 - 3 = 1 For n = 2, 6 - 9 = -3 (which is equivalent to 7 because we are considering only the last digit) For n = 3, 4 - 7 = -3 (which is equivalent to 7) For n = 4, 6 - 1 = 5 For n = 5, 4 - 3 = 1 (the pattern repeats)

So, the last digit (x) of the expression 4n - 3n can be 1, 7, or 5.

Step 3: Count the number of possible values of x.

There are 3 possible values of x: 1, 7, and 5.

So, the number of possible values of x is 3.

This problem has been solved

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